533,884 views
32 votes
32 votes
Can somebody help me solve question 1 ? 2. The next steps are going to be procedural:(a) subtract the term a0 from both sides of the equation.(b) multiply both sides by q^n to rid the equation of fractions.(c) now, factor out the greatest common factor.d) since both sides of the equation must be integers , what does this mean regarding the possibility of p being a factor of a0 ? Keep in mind that p and q have no factors in common .

Can somebody help me solve question 1 ? 2. The next steps are going to be procedural-example-1
User Alexander Luna
by
3.3k points

1 Answer

11 votes
11 votes

Given the Polynomial Function in the form:


f(x)=a_nx^n+a_(n-1)x^(n-1)_{}+\cdots+a_2x^2+a_1x+a_0

1. This is the Leading Coefficient:


a_n

And this is the Constant term:


a_0

The Rational Root Theorem states that a Rational Root of the function has this form:


(p)/(q)

Where "p" is a factor of the Constant Term, and "q" is a factor of the Leading Coefficient.

In this case, you know that this is a root of the function f(x):


x=(p)/(q)

That indicates that when you substitute that root into the function, you will obtain that:


f(x)=0

Because, by definition, the roots of a function are the x-intercept, and the y-value is zero when the function intersects the x-axis.

Then, in order to solve the exercise, you must substitute the root into the function and make the function equal to zero:


a_n((p)/(q))^n+a_(n-1)((p)/(q))^(n-1)_{}+\cdots+a_2((p)/(q))^2+a_1((p)/(q))+a_0=0

2. The exercise indicates that you must;

(a) Subtract the Constant Term from both sides of the equation:


\begin{gathered} a_n((p)/(q))^n+a_(n-1)((p)/(q))^(n-1)_{}+\cdots+a_2((p)/(q))^2+a_1((p)/(q))+a_0-(a_0)=0-(a_0) \\ \\ a_n((p)/(q))^n+a_(n-1)((p)/(q))^(n-1)_{}+\cdots+a_2((p)/(q))^2+a_1((p)/(q))=-a_0 \end{gathered}

(b) Now you have to multiply both sides of the equation by:


q^n

But first you can rewrite the equation applying the Power of Quotients Property, which states that:


((a)/(b))^m=\frac{a^m^{}}{b^m}^{}

Then:


a_n\cdot\frac{p^n^{}}{q^n}^{}+a_(n-1)\cdot(p^(n-1))/(q^(n-1))^{}_{}+\cdots+a_2\cdot(p^2)/(q^2)+a_1\cdot(p)/(q)=-a_0

Therefore, multiplying both sides by the term shown before:


\begin{gathered} (q^n)(a_n\cdot(p^n)/(q^n)^{})+(a_(n-1)\cdot(p^(n-1))/(q^(n-1))^{}_{})(q^n)+\cdots+(a_2\cdot(p^2)/(q^2))(q^n)+(a_1\cdot(p)/(q))(q^n)=(-a_0)(q^n) \\ \\ a_n\cdot(p^nq^n)/(q^n)^{}+a_(n-1)\cdot(p^(n-1)q^n)/(q^(n-1))^{}_{}+\cdots+a_2\cdot(p^2q^n)/(q^2)+a_1\cdot(pq^n)/(q)=-a_0q^n \end{gathered}

Now that you set up the multiplication you need to apply the Quotient of Powers Property to simplify the terms on the left side. This states that:


(b^m)/(b^z)=b^(m-z)

Therefore, since in the numerator and in the denominator you have the same base "q", you can subtract the exponents:


a_n\cdot p^nq^(n-n)+a_(n-1)\cdot p^(n-1)q^(n-(n-1))+\cdots+a_2\cdot p^2q^(n-2)+a_1\cdot pq^(n-1)=-a_0q^n
a_n\cdot p^n+a_(n-1)\cdot p^(n-1)q^{}+\cdots+a_2\cdot p^2q^(n-2)+a_1\cdot pq^(n-1)=-a_0q^n

(c) The Greatest Common Factor (GCF) is the largest factor all the terms have in common. In this case, there every term in the equation has a factor "p". Therefore:


GCF=p

Then, you can factor it out:


p(a_n\cdot p^(n-1)+a_(n-1)\cdot p^(n-2)q+\ldots a_2\cdot p^{}q^(n-2)+a_(1\cdot)q^(n-1))=-a_0q^n

Remember that according to the Product of Powers Property, when two powers with equal bases are multiplying, you have to add the exponents. You need to keep in mind this property to factor it out.

(d) Knowing that both sides of the equation must be integers and knowing that "p" and "q" have no factors in common, you can conclude that "p" can only divide the Constant Term. Then:


a_n\cdot p^(n-1)+a_(n-1)\cdot p^(n-2)q+\ldots a_2\cdot p^{}q^(n-2)+a_(1\cdot)q^(n-1)=(-a_0)/(p)q^n

That means that "p" is a factor of the Constant Term.

Hence, the answers are:

1.


a_n((p)/(q))^n+a_(n-1)((p)/(q))^(n-1)_{}+\cdots+a_2((p)/(q))^2+a_1((p)/(q))+a_0=0

2. (a)


a_n((p)/(q))^n+a_(n-1)((p)/(q))^(n-1)_{}+\cdots+a_2((p)/(q))^2+a_1((p)/(q))=-a_0

(b)


a_n\cdot p^n+a_(n-1)\cdot p^(n-1)q^{}+\cdots+a_2\cdot p^2q^(n-2)+a_1\cdot pq^(n-1)=-a_0q^n

(c)


p(a_n\cdot p^(n-1)+a_(n-1)\cdot p^(n-2)q+\ldots a_2\cdot p^{}q^(n-2)+a_(1\cdot)q^(n-1))=-a_0q^n

(d) It means that "p" is a factor of the Constant Term.

User MariaL
by
3.2k points